p (i.e., the set of all nonnegative real numbers). Similarly, Z will denote the set of all

Transcription

1 th Prelmnry E 689 Lecture Notes by B. Yo 0. Prelmnry Notton themtcl Prelmnres It s ssumed tht the reder s fmlr wth the noton of set nd ts elementry oertons, nd wth some bsc logc oertors, e.g. x A : x s n element of the set A x A : x does not belong to A B A : B s subset of A B A : ntersecton of B nd A B A : unon of B nd A b : s true mles tht b s true (b not true mles s not true) b : s true ff (f nd only f) b s true : symbol for ll Throughout the course, R wll denote the set of ll rel numbers; C the set of ll comlex numbers nd R+ = k x R x 0 (.e., the set of ll nonnegtve rel numbers). Smlrly, Z wll denote the set of ll ntegers nd Z + the set of ll nonnegtve ntegers. 0.. Functons Gven two sets X nd Y, we denote functon f by f : X Y to men tht, for every x X, f ssgns one nd only one element f (x) Y. X s clled the domn of f nd we sy tht f ms X to Y. We defne f (X) = {f (x) x X } s the rnge of f. At tmes t s convenent to defne functon exlctly, for exmle t cos(t) mens the functon tht ms t to cos(t). f : X Y s onto f F (X) = Y. f : X Y s one-to-one f f (x ) = f (x ) x = x (or equvlently, x x f (x ) f (x )). If f s one-to-one, then t hs n nverse tht ms f (X) X, whch s normlly reresented s f -.

2 th Prelmnry E 689 Lecture Notes by B. Yo 0. Vector Sces 0.. Algebrc Asects Defnton A feld F s set of elements clled sclrs together wth two bnry oertons, ddton (+) nd multlcton ( ) such tht for ll α, β, γ F the followng roertes hold: () Closure. α β F, α +β F (b) Commuttvty. α β = β α, α + β = β + α (c) Assoctvty. α + (β + γ) = (α + β) + γ, α (β γ) = (α β) γ (d) Dstrbuton. α (β + γ) = (α β) + (α γ) (e) Identty. There exsts n ddtve dentty 0 F nd multlctve dentty F such tht α + 0 = α, α = α (f) Inverses. For ll α F there exsts n ddtve nverse -α F such tht α + (-α) = 0. For ll α F, α 0 nd multlctve nverse α - F such tht α α - = Exmles The followng re exmles of felds: - R = the set of rel numbers - C = the set of comlex numbers - Q = the set of rtonl numbers - R(s) = the set of rtonl functons n s wth rel coeffcents These re not felds: - R[s] = the set of olynomls n s wth rel coeffcents. Why? - R x the set of rel mtrces. Why? Defnton A vector sce (V, F) s set of vectors V together wth feld F nd two oertons vector-vector ddton (+) nd vector-sclr multlcton (o) such tht for ll α, β F nd ll v, v, v 3 V, the followng roertes hold: () Closure. v + v V, α o v V (b) Commuttvty. v + v = v + v (c) Assoctvty. (v + v ) + v 3 = v + (v + v 3 ) (d) Dstrbuton. α o (β o v ) = (α β) o v, α o (v + v ) = α o v + α o v (e) Addtve Identty. There exsts vector 0 V such tht v + 0 = v for ll v V (f) Addtve Inverse. For ll v V, there exsts (-v) V such tht v + (-v) = 0 We shll henceforth suress the cumbersome notton, o s the rorte cton wll be cler from context. Also, we shll often refer to vector sce V wthout exlct reference to the bse feld F (whch wll exclusvely be R or C). We should however cuton the reder tht dfferent choces of the bse feld F result n fundmentlly dfferent vector sces (see exmle below).

3 th Prelmnry E 689 Lecture Notes by B. Yo Exmles The followng re exmles of vector sces: - (R, R), (C, C) wth ddton nd multlcton s defned n the feld. Any feld s vector sce over tself. - (R n, R), (C n, C) wth comonent-wse ddton nd sclr multlcton. - (R[s], R) wth forml ddton nd sclr multlcton of olynomls. - (C, R) s vector sce. Note tht ths vector sce s fundmentlly dfferent from (C, C). Why? - The sce of nfnte sequences of rel numbers x = (x, x,...) wth x R on the feld R. - C[, b] = {f : [, b] R,, f s contnuous } (.e., the set of ll contnuous functons whch m the ntervl [, b] R to R) s vector sce over R wth ontwse ddton nd multlcton. - The Lebesgue sces L [,b], < defned s L [, b] f :[, b] R, f( t) dt zb = < { } re vector sces over R wth ontwse ddton nd multlcton. (We wll lter tlk more bout L vector sces). (R, C) s not vector sce wth the usul comlex rthmetc. Why? Defntons A set (ossbly nfnte) S = {v : I} of vectors from V s clled lnerly deendent f there exst sclrs α, not ll zero nd only fntely mny α beng nonzero such tht = 0 α v I otherwse, the set of vectors S s sd to be lnerly ndeendent. The dmenson of vector sce V s the mxml number of lnerly ndeendent vectors n V. A set B of vectors n V s clled bss for V f every vector n V cn be unquely exressed s fnte lner combnton of vectors n B. Bss re not unque. Exmles -- In the vector sce (R, R), v, v, v = L N Q P = L N Q P = L N Q P s set of lnerly deendent vectors becuse v + v + v 3 = v L N = s + s +, v = L N s s + + 4s+ 3 s + 3 3

4 th Prelmnry E 689 Lecture Notes by B. Yo re lnerly deendent n (R (s), R (s)), but lnerly ndeendent n (R (s), R). Why? -- The set of vectors S = (, t, t, ) re lnerly ndeendent n C [0,]. -- The dmenson of (R n, R) s n nd the set RL S T N P Q L0 P N Q L0 U PV N QW 0 0 B=,,, = e, e, e n P P P 0 0 qulfes s bss for ths vector sce. -- The dmenson of (C n, R) s n. Exhbt bss for ths sce. l Theorem Let V be n n-dmensonl vector sce nd let B be collecton of vectors drwn from V. then, B s bss f nd only f B contns n lnerly ndeendent vectors. Defnton Let V be vector sce. A subset W V s clled subsce f W s tself vector sce. Let S = {v : I} be set of vectors drwn from V. The sn of S s the set of ll fnte lner combntons of vectors n S. We wll denote ths set SP(S). Notce tht S s bss of V f S s lnerly ndeendent set, nd SP(S) = V. Theorem A set W V s subsce f nd only f t s closed under vector ddton nd sclr multlcton,.e. α w W, w + w W for ll α F, w, w W. It s cler tht the zero vector must le n every (nonemty) subsce. Usng the bove result, t s esy to verfy tht gven set of vectors S, SP{S} s subsce. q 0.. Normed Vector Sces In the sequel, we consder vector sces over the feld C of comlex numbers or the feld R of rel numbers. Defnton Let V be vector sce. A norm on V s functon : V R such tht () v 0nd v = 0 v= 0. (b) α v = α v for ll α F, v V. (c) v + v v + v. (Trngle nequlty) A vector sce on whch norm hs been defned s clled normed sce. 4

5 th Prelmnry E 689 Lecture Notes by B. Yo Exmles The followng re exmles of norms. () n R n or C n : n = n = e = j / (Euclden norm) n = < = -- v = v -- v v e j /, (-norm) -- v v -- v = mx v (su norm) where L N v v v = v n () n the sce of nfnte sequences of rel numbers x = (x, x, ) wth x R on the feld R. Frequently used norms on the subsces l, l nd l re resectvely: -- x = x = -- x x, = -- x = su x / = F H G I K J < () n the functon sces L [,b], L [,b], L [,b]: b -- f z = f() t dt -- f z b / = f t dt e () j < -- f = su f( t) t [, b], By su we men the essentl suremum,.e., su f( t) = nf : f( t) < lmost everywhere t [, b] (excet on set of mesure zero). k 5

6 th Prelmnry E 689 Lecture Notes by B. Yo (v) n the sce of mtrces R n n or C n n : -- The nduced -norm -- ther nduced norms -- The Frobenus norm x T = su = λmx( ) = σ mx( ) x 0 x x = su = mx x 0 x x = su = mx x 0 x j j L T = m trce N Q P = ( ) F j, j, n / where m,j s the, j-th element of the mtrx. m m j j (row sum) (column sum) / 0..3 Equvlent Norms Defnton Let V be vector sce on C. Let nd b be two norms on V. The norms nd b re sd to be equvlent f there exst two ostve numbers m l nd m u such tht ml v v m b u v, for ll v V. Theorem All norms on fnte dmensonl sces (e.g., R n, C n ) re equvlent. Exmle n R n : v v n v v v n v n v v v 0..4 Reltons Between Normed Sces Norms n nfnte dmensonl sces re generlly not equvlent. However, severl mortnt reltons cn be estblshed. Theorem n the normed sces l, l nd l,, whch re subsces of the set of nfnte sequences of rel numbers x = (x, x, ) wth x R on the feld R, Proof: l l l, < <. 6

7 th Prelmnry E 689 Lecture Notes by B. Yo Consder ny nteger [, ]. If x l, then [, ]. For ny nteger N 0 nd ny nteger, N N = = x x x F H G I K J d. = x <. Hence, x l nd l l for Hence, s N, f x l, x x <. Thus, l l. Fct: If f : R + R nd f L L (.e., f belongs to both L nd L ), then f L for [, ]. Proof: Defne the set I = {t f(t) }. Snce f L, the z Lebesgue mesure of the set I s fnte. Ths fct, together wth the fct tht f L mles tht f( t) dt <. I Defnng now the comlement of I, I c _ {t f ( t ) < }, zc I zc I f() t dt f() t dt <, for ll [, ]. The concluson follows from these two observtons Inner Product Sces Defnton Let V be vector sce on C. An nner roduct on V s functon <, > : V V C such tht () < v, w > = < wv, > (b) < v, α w > = α < v, w>. (c) < v, w + w > = < v, w > + < v, w >. (d) < v, v > 0, < v, v > = 0 v = 0. where α denotes the comlex conjugte of α C. A vector sce on whch n nner roduct hs been defned s clled n nner roduct sce. Exmles The followng re exmles of nner roducts. -- In R n, < v, w > = v T w -- In C n, < v, w > = v * w where v * denotes the comlex conjugte trnsose of v C n,.e., v * = v T, -- In L [, b], z b < f, g > = f () t g() t dt 7

8 th Prelmnry E 689 Lecture Notes by B. Yo Theorem Let V be n nner roduct sce. Then / v = < v, v > qulfes s norm on V. The norm defned bove s sd to be nduced by the nner roduct. In n nner roduct sce, ths s the nturl norm to use. Cuchy Schwrtz Inequlty Let V be n nner roduct sce. Then / / < vw, > v w An mmedte consequence of the Cuchy-Schwrtz nequlty led to the nner roduct sce L [, b] s zb zb / zb / f () t g() t dt f () t dt g () t dt L N L N Defnton In n nner roduct sce V, two vectors v, w re sd to be orthogonl f < v, w > = 0. Ths s often wrtten s v w. Further, v s orthogonl to the set of vectors S f v w for ll w S. Ths s often wrtten s v S. A set of vectors S s clled orthogonl f v w for ll v w, v, w S nd s clled orthonorml f n ddton v = for ll v S. We now rovde smle geometrc nterretton of the nner roduct. Let V be n nner-roduct sce nd fx v V. Let b V be unt vector nd consder the subsce S = Sn{b}. Defne s = < v, b > s We wsh to fnd n otml roxmton for v from the subsce S. ore recsely we wsh to solve the followng roblem: mn v s s S We hve the followng result: Lemm The vector s bove s the otml roxmton n S of v. Thus < v, b > my be regrded s the length of the rojecton of v on S. 8

9 th Prelmnry E 689 Lecture Notes by B. Yo 0.3 Hermtn nd Postve Defnte trces Defnton A mtrx U C n n s clled untry f U U = I = UU. A rel untry mtrx s clled n orthogonl mtrx. Lemm Let U C n n be untry nd consder the Hlbert sce C n equed wth the usul nner roduct. Then, () The columns of U form n orthonorml bss of C n. (b) Ux = x (c) < Ux, Uy > = < x, y > (d) U - = U *. Rotton mtrces re untry. Defnton A mtrx H C n n s clled Hermtn f H = H *. Symmetrc mtrces re n rtculr Hermtn. We wll now rove severl results regrdng Hermtn mtrces. These results lso hold lmost verbtm for symmetrc mtrces. Theorem The egenvlues of Hermtn mtrx H re ll rel. Theorem A Hermtn mtrx H hs full set of egenvectors. oreover, these egenvectors form n orthogonl set. As consequence, Hermtn mtrces cn be dgonlzed by untry trnsformtons,.e., there exsts untry mtrx U such tht H = UDU * where D s dgonl mtrx whose entres re the (rel) egenvlues of H. Theorem Let H C n n be Hermtn. Then, () H Hv = su = λ mx( H) v 0 v vhv * * (b) nf mn( H) v = λ or λmn( Hvv ) vhv λmx( Hvv ), v 0 v v Defnton A mtrx P C n n s clled ostve-defnte nd wrtten s P > 0 f P s Hermtn nd further, vpv > 0, forll 0 v C n A mtrx P C n n s clled ostve-sem-defnte nd wrtten s P f P s Hermtn nd further, vpv 0, forll v C n Anlogous re the notons of negtve nd negtve-sem-defnte mtrces. 9

10 th Prelmnry E 689 Lecture Notes by B. Yo Theorem Let P C n n be Hermtn. The followng re equvlent. () P > 0 (b) All the egenvlues of P re ostve. (c) All the ledng rncl mnors of P re ostve. Exmle For the mtrx P = L N the ledng rncl mnors re R L, DetS T N U V W nd Det {P}. Theorem Let P C n n be Hermtn. The followng re equvlent. () P 0 (b) All the egenvlues of P re 0. (d) P = N * N, where N s ny mtrx. A rncl mnor test for ostve-sem-defnteness s sgnfcntly more comlcted. Lemm Let 0 < P C n n nd let X C n m. () x = x Px qulfes s norm on C n (b) X PX 0 (c) X PX > 0 f nd only f rnk (X) = m Defnton Let 0 P C n n. We cn then wrte P = UDU where U s untry. Defne the squreroot of P (wrtten sp / ) by / / P = UD U It s evdent tht P / s defned bove s Hermtn, nd moreover P / 0. Furthermore, f P > 0, then P / > 0. 0